Is ${262455}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {262455}= &&{2}\cdot100000+ \\&&{6}\cdot10000+ \\&&{2}\cdot1000+ \\&&{4}\cdot100+ \\&&{5}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {262455}= &&{2}(99999+1)+ \\&&{6}(9999+1)+ \\&&{2}(999+1)+ \\&&{4}(99+1)+ \\&&{5}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {262455}= &&\gray{2\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {2}+{6}+{2}+{4}+{5}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${262455}$ is divisible by $9$ if ${ 2}+{6}+{2}+{4}+{5}+{5}$ is divisible by $9$ Add the digits of ${262455}$ $ {2}+{6}+{2}+{4}+{5}+{5} = {24} $ If ${24}$ is divisible by $9$ , then ${262455}$ must also be divisible by $9$ ${24}$ is not divisible by $9$, therefore ${262455}$ must not be divisible by $9$.